PAT 甲级 1153 Decode Registration Card of PAT

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最新推荐文章于 2023-05-09 15:27:49 发布

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分类专栏： PAT甲级 PAT甲级/乙级机试经验文章标签： PAT 甲级 Decode Registration Card of PA 1153

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本文介绍了一个竞赛题目，要求解析PAT注册卡并根据不同的查询条件统计参赛者成绩。注册卡由四个部分组成：级别、考点编号、考试日期和考生编号。输入包括考生的注册卡和分数，以及各种查询请求。

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1153 Decode Registration Card of PAT （25 point(s)）

A registration card number of PAT consists of 4 parts:

the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic;
the 2nd - 4th digits are the test site number, ranged from 101 to 999;
the 5th - 10th digits give the test date, in the form of yymmdd;
finally the 11th - 13th digits are the testee's number, ranged from 000 to 999.

Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤104) and M (≤100), the numbers of cards and the queries, respectively.

Then N lines follow, each gives a card number and the owner's score (integer in [0,100]), separated by a space.

After the info of testees, there are M lines, each gives a query in the format Type Term, where

Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term will be the letter which specifies the level;
Type being 2 means to output the total number of testees together with their total scores in a given site. The corresponding Term will then be the site number;
Type being 3 means to output the total number of testees of every site for a given test date. The corresponding Term will then be the date, given in the same format as in the registration card.

Output Specification:

For each query, first print in a line Case #: input, where # is the index of the query case, starting from 1; and input is a copy of the corresponding input query. Then output as requested:

for a type 1 query, the output format is the same as in input, that is, CardNumber Score. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed);
for a type 2 query, output in the format Nt Ns where Nt is the total number of testees and Ns is their total score;
for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt's, or in increasing order of site numbers if there is a tie of Nt.

If the result of a query is empty, simply print NA.

Sample Input:

8 4
B123180908127 99
B102180908003 86
A112180318002 98
T107150310127 62
A107180908108 100
T123180908010 78
B112160918035 88
A107180908021 98
1 A
2 107
3 180908
2 999

Sample Output:

Case 1: 1 A
A107180908108 100
A107180908021 98
A112180318002 98
Case 2: 2 107
3 260
Case 3: 3 180908
107 2
123 2
102 1
Case 4: 2 999
NA

经验总结：

emmmm 第一遍直接AC，这一次就。。。一个段错误，一个超内存，很尴尬- -|||所以考试能不能AC，还是看状态的= =，这一题，实际上难的点，就在第三种类型的存储上面，可以肯定要用映射（map或者unordered_map），至于map的类型是什么，这就是一个问题，map<string,map<int,int> >就可以了，因为只要存储考场号以及人数，但是，这种情况没法在查询前进行排序，当然，也没必要进行，在查询时，针对查询的考场，建立一个vector<pair<int,int> > 收集其中的考场号和人数，然后再进行排序，最后输出就行啦~

AC代码

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <unordered_map>
using namespace std;
int m,n;
struct student
{
	int score;
	char id[15];
	student(char a[],int s)
	{
		strcpy(id,a);
		score=s;
	}
};
struct school
{
	int tscore,tnum;
	school():tscore(0),tnum(0){}
}sch[1000];

bool cmp1(student a,student b)
{
	if(a.score!=b.score)
		return a.score>b.score;
	return strcmp(a.id,b.id)<0;
}

bool cmp2(pair<int,int> a,pair<int,int> b)
{
	if(a.second!=b.second)
		return a.second>b.second;
	return a.first<b.first;
}
unordered_map<int,int> rmp;
unordered_map<string,unordered_map<int,int> > mp;
vector<student> stu[3];
int main()
{
	char temp[15],sno[8];
	int score,no,f;
	rmp['B']=0;rmp['A']=1;rmp['T']=2;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;++i)
	{
		scanf("%s%d",temp,&score);
		stu[rmp[temp[0]]].push_back(student(temp,score));
		for(int j=1;j<4;++j)
			sno[j-1]=temp[j];
		sno[3]='\0';
		sscanf(sno,"%d",&no);
		++sch[no].tnum;
		sch[no].tscore+=score;
		for(int j=4;j<10;++j)
			sno[j-4]=temp[j];
		sno[6]='\0';
		++mp[sno][no];
	}
	for(int i=0;i<3;++i)
		sort(stu[i].begin(),stu[i].end(),cmp1);
	for(int i=1;i<=m;++i)
	{
		scanf("%d %s",&f,&temp);
		printf("Case %d: %d %s\n",i,f,temp);
		int level=rmp[temp[0]];
		if(f==1)
			if(stu[level].size()==0)
				printf("NA\n");
			else
				for(int j=0;j<stu[level].size();++j)
					printf("%s %d\n",stu[level][j].id,stu[level][j].score);
		else if(f==2)
		{
			int room;
			sscanf(temp,"%d",&room);
			if(sch[room].tnum==0)
				printf("NA\n");
			else
				printf("%d %d\n",sch[room].tnum,sch[room].tscore);
		}
		else
		{
			if(mp[temp].size()==0)
				printf("NA\n");
			else
			{
				vector<pair<int,int> > x;
				for(unordered_map<int,int>::iterator it=mp[temp].begin();it!=mp[temp].end();++it)
					x.push_back(make_pair(it->first,it->second));
				sort(x.begin(),x.end(),cmp2);
				for(int j=0;j<x.size();++j)
					printf("%d %d\n",x[j].first,x[j].second);
			}
		}
	}
	return 0;
}